HDU 5878 I Count Two Three （DFS预处理 + 二分）

大体题意：

给你一个数字n 找出一个大于n 的最小数字，满足它的素因子只有2,3,5,7组成！

思路：

直接利用set 进行dfs打表，set 中已经存在了就不用在dfs了，打完表直接二分查找即可！

详细见代码：

#include <cstdio>#include <cstring>#include <algorithm>#include <set>using namespace std;typedef long long ll;const ll maxn = (ll)(1e9+10)*8ll;set<ll>s;ll a[5] = {2,3,5,7};void dfs(ll v){if (v > maxn)return;if (s.count(v))return;s.insert(v);for (int i = 0; i < 4; ++i){dfs(v*a[i]);}}int main(){dfs(1ll);int T;scanf("%d",&T);while(T--){ll n;scanf("%I64d",&n);printf("%I64d\n",*(s.lower_bound(n)));}return 0;}

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 189    Accepted Submission(s): 108

Problem Description

I will show you the most popular board game in the Shanghai Ingress Resistance Team.
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.

After the event, we analysed the laws of failed attacks.
It's interesting that the i-th attacks failed if and only if i can be rewritten as the form of 2a3b5c7d which a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form 2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n.

 

Input

The first line of input contains an integer t (1≤t≤500000), the number of test cases. t test cases follow. Each test case provides one integer n (1≤n≤109).

 

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than n.

 

Sample Input

1011113123123412345123456123456712345678123456789

 

Sample Output

11214125125012348123480123480012348000123480000

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 

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